Calculus Examples

${(e^{x} + 1)}^{2}$

Step 1

Write ${(e^{x} + 1)}^{2}$ as a function.

$f (x) = {(e^{x} + 1)}^{2}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int {(e^{x} + 1)}^{2} d x$

Step 4

Simplify.

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Step 4.1

Rewrite ${(e^{x} + 1)}^{2}$ as $(e^{x} + 1) (e^{x} + 1)$ .

$\int (e^{x} + 1) (e^{x} + 1) d x$

Step 4.2

Expand $(e^{x} + 1) (e^{x} + 1)$ using the FOIL Method.

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Step 4.2.1

Apply the distributive property.

$\int e^{x} (e^{x} + 1) + 1 (e^{x} + 1) d x$

Step 4.2.2

Apply the distributive property.

$\int e^{x} e^{x} + e^{x} \cdot 1 + 1 (e^{x} + 1) d x$

Step 4.2.3

Apply the distributive property.

$\int e^{x} e^{x} + e^{x} \cdot 1 + 1 e^{x} + 1 \cdot 1 d x$

Step 4.3

Simplify and combine like terms.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 4.3.1.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int e^{x + x} + e^{x} \cdot 1 + 1 e^{x} + 1 \cdot 1 d x$

Step 4.3.1.1.2

Add $x$ and $x$ .

$\int e^{2 x} + e^{x} \cdot 1 + 1 e^{x} + 1 \cdot 1 d x$

Step 4.3.1.2

Multiply $e^{x}$ by $1$ .

$\int e^{2 x} + e^{x} + 1 e^{x} + 1 \cdot 1 d x$

Step 4.3.1.3

Multiply $e^{x}$ by $1$ .

$\int e^{2 x} + e^{x} + e^{x} + 1 \cdot 1 d x$

Step 4.3.1.4

Multiply $1$ by $1$ .

$\int e^{2 x} + e^{x} + e^{x} + 1 d x$

Step 4.3.2

Add $e^{x}$ and $e^{x}$ .

$\int e^{2 x} + 2 e^{x} + 1 d x$

Step 5

Split the single integral into multiple integrals.

$\int e^{2 x} d x + \int 2 e^{x} d x + \int d x$

Step 6

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = 2 x$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 6.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x]$

Step 6.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1$

Step 6.1.4

Multiply $2$ by $1$ .

$2$

Step 6.2

Rewrite the problem using $u$ and $d u$ .

$\int e^{u} \frac{1}{2} d u + \int 2 e^{x} d x + \int d x$

Step 7

Combine $e^{u}$ and $\frac{1}{2}$ .

$\int \frac{e^{u}}{2} d u + \int 2 e^{x} d x + \int d x$

Step 8

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int e^{u} d u + \int 2 e^{x} d x + \int d x$

Step 9

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$\frac{1}{2} (e^{u} + C) + \int 2 e^{x} d x + \int d x$

Step 10

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$\frac{1}{2} (e^{u} + C) + 2 \int e^{x} d x + \int d x$

Step 11

The integral of $e^{x}$ with respect to $x$ is $e^{x}$ .

$\frac{1}{2} (e^{u} + C) + 2 (e^{x} + C) + \int d x$

Step 12

Apply the constant rule.

$\frac{1}{2} (e^{u} + C) + 2 (e^{x} + C) + x + C$

Step 13

Simplify.

$\frac{1}{2} e^{u} + 2 e^{x} + x + C$

Step 14

Replace all occurrences of $u$ with $2 x$ .

$\frac{1}{2} e^{2 x} + 2 e^{x} + x + C$

Step 15

The answer is the antiderivative of the function $f (x) = {(e^{x} + 1)}^{2}$ .

$F (x) =$ $\frac{1}{2} e^{2 x} + 2 e^{x} + x + C$